$Find $\lim \limits_{x \rightarrow \infty} \frac{x e^{-x}+1}{1+e^{-x}}$$

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$Find $\lim \limits_{x \rightarrow \infty} \frac{x e^{-x}+1}{1+e^{-x}}$$

I am stuck on this question. Please help me find the limit.

Precalculus Limits Calculus

Roman Davis

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We have
$$\lim_{x \rightarrow \infty} \frac{xe^{-x}+1}{1+e^{-x}}=\lim_{x \rightarrow \infty} \frac{e^{x}(xe^{-x}+1)}{e^x(1+e^{-x})}=\lim_{x \rightarrow \infty} \frac{x+e^{x}}{e^x+1}$$
Using the L'Hospital's rule we get
$$\lim_{x \rightarrow \infty} \frac{x+e^{x}}{e^x+1}=\lim_{x \rightarrow \infty} \frac{1+e^{x}}{e^x}=\lim_{x \rightarrow \infty} \frac{e^{x}}{e^x}=1.$$
We have used the L'Hospital's rule twice to get the last two equalities.

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$Find $\lim \limits_{x \rightarrow \infty} \frac{x e^{-x}+1}{1+e^{-x}}$$

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